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The nonlinear boundary layer to the Boltzmann equation with mixed boundary conditions for hard potentials
J. Math. Anal. Appl. 375 (2011) 725–737
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The nonlinear boundary layer to the Boltzmann equation with mixed boundary conditions for hard potentials Jie Sun a,b,c,∗ , Qianzhu Tian a,b,c,1 a b c
Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China Department of Mathematics, City University of Hong Kong, Kowloon, Hong Kong Joint Advanced Research Center of University of Science and Technology of China and City University of Hong Kong, Suzhou, Jiangsu 215123, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 18 April 2010 Available online 13 October 2010 Submitted by Steven G. Krantz
In this paper, the existence of boundary layer solutions to the Boltzmann equation for hard potential with mixed boundary condition, i.e., a linear combination of Dirichlet boundary condition and diffuse reflection boundary condition at the wall, is considered. The boundary condition is imposed on the incoming particles, and the solution is supposed to approach to a global Maxwellian in the far field. As for the problem with Dirichlet boundary condition (Chen et al., 2004 [5]), the existence of a solution highly depends on the Mach number of the far field Maxwellian. Furthermore, an implicit solvability condition on the boundary data which shows the codimension of the boundary data is related to the number of the positive characteristic speeds is also given. © 2010 Elsevier Inc. All rights reserved.
Keywords: Boltzmann equation Boundary layer Hard potential Mixed boundary condition
1. Introduction In our paper, we consider the boundary layer problem of the Boltzmann equation for hard potential with angular cutoff which arises in the physical condensation–evaporation problem. We will study the stationary problem in a half-space with a mixed boundary condition, i.e., a linear combination of Dirichlet boundary condition and diffuse reflection boundary condition and convergence to an equilibrium state at infinity. A lot of numerical results about the boundary layer problem have been made by [1,2,11,12]. The boundary layer problem for linearized Boltzmann equation has been well-studied in [3,6,9]. The existence of solutions was solved in [8] on the nonlinear problem with specular reflection boundary condition. With respect to the nonlinear problem with Dirichlet boundary conditions, existence and stability of the boundary layer solutions are given by [5,16,18] and [17,19,20] respectively. Recently, the problems with mixed boundary conditions for hard sphere case were considered in [13,14]. In our paper, we will work on the boundary layer problem with a mixed boundary condition by using the method developed in [5] for hard potential case. The boundary layer problem we consider is as follows,
⎧ x > 0, ξ ∈ R 3 , ξ1 F x = Q ( F , F ), ⎪ ⎪ ⎪ ⎨ ξ F 0 , ξ d ξ , ξ1 > 0 , F |x=0 = F 0 (ξ ) + sM ω 1 ⎪ ⎪ ξ1 <0 ⎪ ⎩ F → M ∞ (x → ∞), ξ ∈ R3.
*
(1.1)
Corresponding author at: Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China. E-mail addresses: [email protected] (J. Sun), [email protected] (Q. Tian). 1 The author’s research was supported both by the National Natural Science Foundation of China Grant No. 10871187, and the Natural Science Foundation of Anhui Grant No. 070416228. 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.09.067
©
2010 Elsevier Inc. All rights reserved.
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J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
As usual, the collision operator takes the form
F ξ F ξ∗ − F (ξ ) F (ξ∗ ) q( v , θ) dξ∗ dω,
Q (F , F ) = R3×S2
where ξ and ξ∗ are velocities of gas particles before collision, while ξ = ξ − ω(ξ − ξ∗ ) · ω and ξ∗ = ξ∗ + ω(ξ − ξ∗ ) · ω are velocities after collision. Here v = ξ − ξ∗ , ω ∈ S 2 and cos θ = v| v·ω| . This is a stationary problem with mixed boundary condition at x = 0 for incoming particles, where M ∞ (ξ ) = M [ρ∞ ,u ∞ , T ∞ ] (ξ ) and M ω (ξ ) = M [1,0, T ω ] (ξ ) are Maxwellians. As to the collision kernel q( v , θ), we have the following assumption for the hard potential with angular cutoff, cf. [15]. Assumption. There exist 0 δ < 1 and a positive constant c such that
0 q( v , θ) c | v | + | v |−δ | cos θ|;
(1.2)
there is a positive constant c,
e−
|ξ∗ |2 2
q( v , θ) dξ∗ dω
R3×S2
e −|ξ∗ | q( v , θ) dξ∗ dω 2
−1
c.
R3×S2
In this paper, the gas constant R is normalized to 1. Then a Maxwellian with the mass density temperature T is given by
M=
(1.3)
ρ (2π T )
3 2
e−
|ξ −u |2 2π T
ρ , flow velocity u and
.
It is well known that Q ( M ) = 0. Furthermore, the collision operator has exactly five collision invariants
φ 0 = 1,
φi = ξ
φ4 = |ξ |2
(i = 1, 2, 3),
satisfying
φ i , Q ( F ) = 0,
i = 0, 1 , . . . , 4 ,
for any distribution function F (see [7]), here , is an inner product of L 2ξ . Without loss of generality, we can assume u ∞,2 and u ∞,3 of the Maxwellian in the far field are zeroes. Then the sound speed and Mach number of the far field Maxwellian can be defined
c∞ =
5 3
M∞ =
T ∞,
u ∞,1 c∞
(1.4)
, 1
2 see [4]. We will consider the boundary layers near the global Maxwellian M ∞ at infinity. Let F = M ∞ + M ∞ f , then Eq. (1.1) will be reformulated into
⎧ x > 0, ξ ∈ R 3 , ⎪ ⎪ ξ1 f x − L f = Γ ( f , f ), ⎪ ⎪ 1 1 1 ⎨ − − 2 ξ M ∞ + M ∞ f |x=0 = M ∞2 ( F 0 − M ∞ ) + sM ∞2 M ω f 0 , ξ d ξ , ξ1 > 0 , 1 ⎪ ⎪ ξ1 <0 ⎪ ⎪ ⎩ f → 0 (x → ∞), ξ ∈ R 3,
(1.5)
where 1 − 1 12 2 Γ ( f , f ) = M ∞2 Q M ∞ f , M∞ f
and −1
1
1
2 2 L f = M ∞2 Q M ∞ , M ∞ f + Q M∞ f , M∞ .
In the following, we will introduce two weight functions. The first one is
−β
W β (ξ ) = 1 + |ξ |
M 1/2 [1, u ∞ , T ∞ ](ξ )
(1.6)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
with β ∈ R. The second one is
δx + l (1 + |ξ − u ∞ |)3−β1
δx + l δx + l 2 + + 3|ξ − u ∞ | η (1 + |ξ − u ∞ |)1−β1 (1 + |ξ − u ∞ |)3−β1 2
727
σ (x, ξ ) = 5(δ x + l) 3−β1 1 − η
where x 0 and
η (s) =
(1.7)
η : (0, ∞) → R is a smooth non-increasing function which has the following properties: s 1, s 2,
1, 0,
and 0 η(s) 1. After the introduction of the above notations, we state our main result in this paper. Theorem 1.1. Suppose M∞ = 0, ±1 and β > 5/2. If s is small enough and T ω < 2T ∞ , then there exist positive numbers , and a C 1 map
+ Ψ : L 2 R 3+ , e σ (0,ξ ) ξ1 dξ → R n ,
Ψ (0) = 0,
1 , 2 , (1.8)
such that the following holds. (i) For any F 0 satisfying
σ (0,ξ ) 1/2 1 W β , e F − M + sM M 0 , ξ d ξ ξ 0 ∞ ω ∞ 1
ξ ∈ R3+ ,
(1.9)
ξ <0 1
Eq. (1.1) admits a unique solution F in the class
σ (x,ξ ) −1/2 e M ∞ ( F − M ∞ )σ β 2 ,
(1.10)
when F 0 satisfies
1/2 −1/2 ξ M ∞ 0, ξ dξ =0 Ψ M∞ F 0 − M ∞ + sM ω 1
(1.11)
ξ <0 1
where l > 0 is large and δ > 0 is small in σ (x, ξ ) and · σ β is defined in (4.1). (ii) The set of F 0 satisfying (1.9) and (1.11) forms a local C 1 manifold of co-dimension n+ . Remark 1.2. The number n+ of solvability conditions changes with Mach number as
⎧ 0, ⎪ ⎪ ⎪ ⎨ 1, n+ = ⎪ 4, ⎪ ⎪ ⎩ 5,
M∞ ∈ (−∞, −1), M∞ ∈ (−1, 0),
(1.12)
M∞ ∈ (0, 1), M∞ ∈ (0, ∞).
We exclude the cases M∞ = 0, ±1 which are also important physically. 2. Preparations Before we study the equation itself, we will study the properties of the weight functions and the linearized collision operator. 2.1. Weight functions
σ ⎧ c 1 (1 + |ξ − u ∞ |)2 , ⎪ ⎪ ⎨ 2 σ (x, ξ ) = c2 (δ x + l) 3−β1 + c3 (1 + |ξ − u ∞ |)2 , ⎪ ⎪ 2 ⎩ 5(δ x + l) 3−β1 ,
A careful analysis helps us to get the properties of
below:
(x, ξ ) ∈ Ω1 , (x, ξ ) ∈ Ω2 , (x, ξ ) ∈ Ω3 ,
(2.1)
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J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
where
⎧ Ω1 = (x, ξ ): ⎪ ⎪ ⎨ Ω2 = (x, ξ ): ⎪ ⎪ ⎩ Ω3 = (x, ξ ):
3−β1 , δ x + l 1 + |ξ − u ∞ | 3−β1 3−β1 , 1 + |ξ − u ∞ | δ x + l 2 1 + |ξ − u ∞ | 3−β1 δ x + l 2 1 + |ξ − u ∞ | ,
c 1 , c 2 and c 3 are positive functions of (x, ξ ), and c 1 and c 2 + c 3 have uniform lower bounds. With respect to σx , we have the following results:
⎧ δ(1 + |ξ − u ∞ |)−1+β1 , (x, ξ ) ∈ Ω1 , ⎪ ⎪ ⎪ ⎨ 1−β − 3−β1 − 1+β1 1 ), + c 5 (δ x + l) (x, ξ ) ∈ Ω2 , σx = δ(c4 (1 + |ξ − u ∞ |) ⎪ ⎪ 1−β1 ⎪ ⎩ 10δ (δ x + l)− 3−β1 , (x, ξ ) ∈ Ω3 , 3−β1
(2.2)
where c 4 + c 5 has a uniform lower bound. From the above results, we can learn 2
σ (x, ξ ) c6 (δ x + l) 3−β1 ,
1−β
− 3−β1
0 < c 7 min (δ x + l)
1
1−β1 −1+β1 − σx c 8 (δ x + l) 3−β1 , , 1 + |ξ − u ∞ |
β |σx ξ1 | c 1 + |ξ | 1 c ν ,
(2.3)
where c 6 , c 7 and c 8 are positive constants. 2.2. Linearized collision operator It is well known that the linearized collision operator L is of the form:
L f = − f (ξ )
1
2 M ∞ (ξ∗ )q( v , θ) dξ∗ dω − M ∞
1 2 f (ξ∗ ) M ∞ (ξ∗ )q( v , θ) dξ∗ dω
1 12 12 2 + f ξ∗ M ∞ ξ + f ξ M ∞ ξ∗ M ∞ (ξ∗ )q( v , θ) dξ∗ dω = −ν (ξ ) − K 1 + K 2 f (ξ ).
(2.4)
In the sequel, we assume that
β 1
c 1 1 + |ξ |
β 2 ν (ξ ) c 2 1 + |ξ |
for constants 0 < β1 β2 1. K 1 and K 2 are integral compact operators with kernels k1 (ξ, ξ∗ ) and k2 (ξ, ξ∗ ) respectively. Set K = − K 1 + K 2 and k = −k1 + k2 , cf. [10]. The kernel k1 has the following property:
k1 (ξ, ξ∗ ) |ξ − ξ∗ | + |ξ − ξ∗ |
−δ0
|ξ − u ∞ |2 |ξ∗ − u ∞ |2 exp − − 4T ∞
4T ∞
where 0 δ0 < 1. Furthermore, the kernel k2 (ξ, ξ∗ ) satisfies,
k2 (ξ, ξ∗ ) = a(ξ, ξ∗ ) exp −
||ξ − u ∞ |2 − |ξ∗ − u ∞ ||2 |ξ∗ − ξ |2 − 8T ∞ 8T ∞ |ξ∗ − ξ |2
(2.5)
(2.6)
where
a(ξ, ξ∗ ) c |ξ − ξ∗ |−1 .
(2.7)
Then, k has the following properties:
R3
k(ξ, ξ∗ ) 1 + |ξ∗ | −β dξ∗ 1 + |ξ | −β−1 ,
β ∈ R,
k(ξ, ξ∗ )2 dξ c ,
R3
∞ ∞ −∞ −∞
k(ξ, ξ∗ ) dξ2 dξ3 c .
(2.8)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
729
At the end of this section, we will state two lemmas which have been proved in [5]. Lemma 2.1. There is a constant 1 > 0, such that for 0 1 and g ∈ N ⊥ ,
2 2 − g , e |ξ −u ∞ | Le − |ξ −u ∞ | g ν1 ν |ξ | g , g ,
(2.9)
for a positive constant ν1 depending on 1 . Lemma 2.2. There is a constant 2 > 0, such that for 0 2 and g ∈ N ⊥ ,
− g , e σ (x,ξ ) Le − σ (x,ξ ) g ν2 ν |ξ | g , g ,
(2.10)
for a positive constant ν2 depending on 2 . 3. Linear existence In order to solve Eq. (1.1), we will firstly show the existence of a solution to the linearized problem (3.5) with damping term. Define the following orthogonal projections with respect to L:
P 0 : L 2ξ → N ,
P 1 = I − P 0 : L 2ξ → N ⊥
(3.1) 1
2 where N is the null space of the linearized collision operator L and is spanned by the five collision invariants M ∞ φi , i = 0, 1, . . . , 4. N ⊥ denotes the orthogonal complement of N in L 2ξ . Define the operator for macroscopic convection:
A = P 0 ξ1 P 0 .
(3.2)
It is easy to see that A has the eigenvalues
λ1 = u ∞,1 − c ∞ ,
λi = u ∞,1 (i = 2, 3, 4),
λ5 = u ∞,1 + c ∞ ,
(3.3)
on N. n+ happens to be the number of the positive eigenvalues. As in [16], we decompose the operator A into the positive part and the negative part A + , A − , and denote the corresponding projections by P + , P − . Note that if M ∞ = 0, ±1, then 0
+
−
A=A +A ,
+
0
−
P0 = P0 + P0 .
We modify (1.1) and (1.5) by adding a damping term,
and
⎧ ξ1 F x = Q ( F , F ) − γ (δ x + l)−Θ P 0+ ξ1 W 0−1 ( F − M ∞ ), x > 0, ξ ∈ R 3 , ⎪ ⎪ ⎪ ⎪ ⎨ ξ F 0, ξ d ξ , F |x=0 = F 0 (ξ ) + sM ω ξ1 > 0 , 1 ⎪ ⎪ ξ 1 <0 ⎪ ⎪ ⎩ F → M ∞ (x → ∞), ξ ∈ R3,
(3.4)
⎧ x > 0, ξ ∈ R 3 , ξ1 f x − L f = h − γ (δ x + l)−Θ P 0+ ξ1 f , ⎪ ⎪ ⎪ ⎪ 12 ⎨ −1 ξ M ∞ f 0, ξ dξ , ξ1 > 0, f |x=0 = b0 (ξ ) + sM ∞2 M ω 1 ⎪ ⎪ ξ1 <0 ⎪ ⎪ ⎩ f → 0 (x → ∞), ξ ∈ R3,
(3.5)
where −1
−1
b0 (ξ ) = M ∞2 ( F 0 − M ∞ ) + sM ∞2 M ω h = Γ ( f , f ),
Θ=
1 − β1 3 − β1
ξ M ∞ dξ , 1
ξ1 <0
,
and γ is a positive constant to be chosen later. Note that for the case M ∞ < −1, A has no positive eigenvalue. Hence, P 0+ = 0 and the damping term vanishes. Let
f = exp − σ (x, ξ ) g .
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J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
The problem (3.5) turns into
⎧ x > 0, ξ ∈ R 3 , ξ g − σx ξ1 g − L g + D g = h¯ , ⎪ ⎪ ⎨ 1 x 1 1 − 2 ξ M ∞ g |x=0 = a0 (ξ ) + sM ∞2 M ω e σ (0,ξ ) e − σ (0,ξ ) g 0, ξ dξ , ξ1 > 0, 1 ⎪ ⎪ ⎩
(3.6)
ξ 1 <0
where
−1
a0 (ξ ) = M ∞2 e σ (0,ξ ) F 0 − M ∞ + sM ω
ξ M ∞ 0, ξ dξ 1
ξ1 <0
and
h¯ = e σ (x,ξ ) h,
D = γ (δ x + l)−Θ e σ P 0+ ξ1 e − σ .
L = e σ Le − σ ,
In the following, we will mainly consider the linear problem (3.6). Let D ∗ be the adjoint operator of D , and set
ψ = −ξ1 φx − σx ξ1 φ − L − φ + D ∗ φ, φ ∈ C 0∞ , 1 1 V = ψ 0 − 2 φ = φ|x=0 = sM ∞ e − σ ξ >0 |ξ1 | M ∞2 M ω e σ φ 0 dξ for ξ1 < 0. 1
(3.7)
We define
g˜ ψ = (h¯ , φ) +
1 2
ξ1 a 0 , φ 0
+
where (,) denotes the inner product in L 2 ( R + × R 3 ) and f , g ± = ±ξ >0 f g dξ . 1 Consider
ψ(φ), φ =
1 − σx ξ1 φ 2 − φ L φ + φ D φ dξ dx +
2 ξ1 φ 0 d ξ
2
R+ R3
+ s2
ξ1 M ∞ e −2 σ
ξ 1 <0
ξ 1 >0
− 12 ξ M ∞ M ω e σ φ 0 dξ 1
2 dξ.
(3.8)
ξ1 >0
In order to estimate the above term, we show the following lemma first. Lemma 3.1. Let l be sufficiently large. Then there exists a positive constant c, such that when γ = O (1) 1 and γ > ,
|ξ˜ |
3−β1
φ γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ dξ c
φ 2 dξ,
(3.9)
R3
2l
for any φ ∈ N where ξ˜ = ξ − u ∞ . Proof. We set φ =
5
χ j , where χ j is the normalized eigenvector of A corresponding to the eigenvalue λ j . Then γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ = (γ − )λ+j b j χ + − λ−j b j χ − , j j j =1 b j
− where λ+ j (λ j ) denotes the positive (negative) eigenvalue, and Therefore, when l is sufficiently large,
− + − χ+ j (χ j ) denotes the corresponding eigenvector to λ j (λ j ).
5 φ γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ dξ < μ b2j = μ i =1
|ξ˜ |3−β1 2l
φ 2 dξ. R3
Moreover, since λ = 0 and
γ = O (1) , there exists a positive constant c, such that φ γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ dξ c φ 2 dξ.
R3
When l is sufficiently large, we have
R3
μ < 2c . Then the proof is completed. 2
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
731
By using Lemma 3.1, we can estimate the term (ψ(φ), φ) as follows. Lemma 3.2. Let l be sufficiently large, and δ be sufficiently small and l−1 . When s is sufficiently small and T ω < 2T ∞ , there exists a constant c, such that when γ = O (1) 1 and γ > , the following result holds
2 1 2
ψ(φ), φ c (δ x + l)−Θ/2 φ0 + ν 2 φ1 + ξ1 φ 0 , φ 0 + .
Proof. Let φ1 = P 1 φ, and φ0 =
5
φ D φ − ξ1 σx φ02 dξ =
R3
j =1 b j
R3
χ j . We consider the term
R3
(3.10)
(φ D φ − ξ1 σx φ02 ) dξ at first,
φ γ (δ x + l)−Θ P 0+ ξ1 P 0 φ − ξ1 σx φ02 dξ
+
φ D − γ (δ x + l)−Θ P 0+ ξ1 P 0 φ dξ.
(3.11)
R3
Notice that
ξ1 σx φ02 dξ R3
10δ 3 − β1
(δ x + l)−Θ
5 2 φ0 P 0 ξ1 φ0 dξ + c (δ x + l)−Θ exp −c (δ x + l) 3−β1 b2j , j =1
2(1+|ξ˜ |)3−β1 δ x+l
then from Lemma 3.1, we have
φ γ (δ x + l)−Θ P 0+ ξ1 P 0 φ − ξ1 σx φ02 dξ c (δ x + l)−Θ φ0 2L 2 .
(3.12)
ξ
R3
We know that
φ D − γ (δ x + l)−Θ P 0+ ξ1 P 0 φ dξ = γ (δ x + l)−Θ φ, e σ P 0+ ξ1 e − σ − P 0+ ξ1 φ + φ, P 0+ ξ1 φ1 .
(3.13)
R3
As the kernel of (e σ P 0+ ξ1 e − σ − P 0+ ξ1 ) is
χ+ (ξ∗ )χ + (ξ )ξ∗1 e (σ (ξ )−σ (ξ∗ )) − 1 , j j
(3.14)
j
we can get that
c φ D − γ (δ x + l)−Θ P 0+ ξ1 P 0 φ dξ (δ x + l)−Θ φ0 2L 2 + c (δ x + l)−Θ φ1 2L 2 . 2
ξ
ξ
R3
From (2.3), we have the following result
1 2 c φ D φ − ξ1 σx φ 2 dξ (δ x + l)−Θ φ0 2L 2 − c (δ x + l)−Θ ν 2 φ1 L 2 . 2
ξ
ξ
R3
Secondly, similar to the case with Dirichlet boundary condition, we also have
1 2 φ L φ dξ −c 2 (δ x + l)−θ φ0 2L 2 + c ν 2 φ1 L 2 .
−
(3.15)
ξ
ξ
R3
Furthermore, when s is sufficiently small, we can have
1 2
s2 ξ 1 <0
ξ1 M ∞ e −2 σ
0 − 12 ξ φ M ∞ M ω e σ dξ 1
ξ1 >0
2 dξ
1 4
ξ1 φ 0 , φ 0 + .
(3.16)
In conclusion, we can get the following result
2 1 2
ψ(φ), φ c (δ x + l)−Θ/2 φ0 + ν 2 φ1 + ξ1 φ 0 , φ 0 +
for some positive constant c. The proof is ended.
2
(3.17)
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J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
In order to achieve existence of a solution to the linear problem (3.6), we define the following norms:
2 1 Θ
u 1 = (δ x + l)− 2 u 0 + u 1 2 2 , 2 1 Θ
u 2 = (δ x + l) 2 u 0 + u 1 2 2 ,
(3.18)
where u 0 = P 0 u and u 1 = P 1 u. Thus, Lemma 3.2 implies that
1 ψ(φ) c φ 1 + c ξ1 φ 0 , φ 0 2 . 2 +
(3.19)
In the following, we get the key result for this section. Theorem 3.3. If h¯ 2 < ∞, T ω < 2T ∞ , , δ and s are sufficiently small, l is sufficiently large and l−1 , then the linearized problem with damping (3.6) has a unique solution g which satisfies
(δ x + l)− Θ2 g 0 2 + ν 12 g 1 2 + ξ1 g 0 , g 0 c ( , a0 ) ξ1 a0 , a0 + + h¯ 2 . 2 −
(3.20)
Proof. By (3.17), we have
1 | g˜ ψ| = (h¯ , φ) + ξ1 a0 , φ 0 + 2
1 0 ¯ ¯ = (h0 , φ0 ) + (h1 , φ1 ) + ξ1 a0 , φ + 2
1 c ( , a0 ) |ξ1 | 2 a0 + + h¯ 2 ψ 2 .
(3.21)
Then, g˜ can be taken as a continuous functional on the subspace V with respect to the norm · 2 . Hence, from the Hahn– Banach theorem and the Riesz representation theorem, we know there exists an element g in the space L 2 ( R + × R 3 ) such that ( g , ψ) = g˜ ψ . A similar argument to [16] yields that g is the unique solution to Eq. (3.6). Multiplying (3.6) by g and integrating the result over R + × R 3 , similar to the discussion in Lemma 3.2, we can get the estimate (3.20). 2 4. Nonlinear existence In this section, we will prove the existence for the nonlinear problem with damping on the basis of the estimates on the solution to the linearized equation (3.6). We will consider the problem in another space. Define the weighted norm:
1 β
f σ β = σx2 1 + |ξ | f L ∞ = x,ξ
1
sup x∈ R ,ξ ∈ R 3
β σx2 1 + |ξ | f (x, ξ ).
(4.1)
In order to obtain the above · σ β norm of the solution to Eq. (3.6), we will introduce the weighted function w α . Set
wα =
|ξ1 |α , |ξ1 | < 1, |ξ1 | 1.
1,
Rewrite (3.6)
g x = σx g +
1
ξ1
L g +
1 ¯ D h− g=
ξ1
ξ1
σx −
ν (ξ ) ξ1
g+
1 ¯ (h + K¯ g )
ξ1
(4.2)
where
K¯ = e σ K e − σ − γ e σ P 0+ ξ1 e − σ . Let
κ (x, ξ ) =
x
ν (ξ ) − σx + dy . ξ1 0
Therefore, the solution to Eq. (3.6) can be formally written as
g = a˜ + U ( K¯ g + h¯ ),
(4.3)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
733
where
a˜ =
− 2 exp(−κ (x, ξ ))(a0 (ξ ) + sM ∞2 M ω e σ (0,ξ ) ξ <0 |ξ1 | M ∞ e − σ (0,ξ ) g (0, ξ ) dξ ), 1
ξ1 > 0 ,
0,
ξ1 < 0
1
and
x 0
U (h) =
−
1
exp(−(κ (x, ξ ) − κ (τ , ξ ))) ξ1 h(τ , ξ ) dτ ,
ξ1 > 0 ,
1
∞
1
exp(−(κ (x, ξ ) − κ (τ , ξ ))) ξ h(τ , ξ ) dτ ,
x
1
(4.4)
(4.5)
ξ1 < 0 .
The operators K¯ and U have the following properties stated in Lemma 4.1 and Lemma 4.2. Lemma 4.1. When is sufficiently small and l is sufficiently large, K¯ satisfies 1. 2.
1
− 12
1 2
− 12
σx2 K¯ σx σx K¯ σx
is a bounded operator from L 2ξ to itself. is a bounded operator from L 2ξ to L ∞ ξ .
3. K¯ h σ β c h σ (β−1) for β ∈ R . 1
− 12
4. w −α σx2 K¯ σx 1 2
− 12
1
−1
is a bounded operator from L 2ξ to itself, when 0 α < 12 .
5.
1 ∞ σx K¯ σx w −α is a bounded operator from L 2ξ ( L ∞ x ) to L x,ξ , when 0 α < 2 .
6.
1 σx2 K¯ σx 2 w −α is a bounded operator from L 2ξ ( L ∞ x ) to itself, when 0 α < 2 .
Proof. The details of the first five properties can be found in [5]. We will prove the sixth property in detail. 1
As
−1
σx2 (x, ξ )e σ (x, ξ )e − σ (x, ξ∗ )σx 2 (x, ξ∗ ) can be controlled by ec ||ξ˜ |
2
−|ξ∗ |2 | , for any h
∈ L 2ξ ( L ∞ x ), we have
1 12 σ σx e K σx− 2 e − σ w −α h22
Lξ (L∞ x )
1
sup
=
x∈ R 3
c
−1
2
σx2 (x, ξ )e σ (x, ξ )k(ξ, ξ∗ )e− σ (x, ξ∗ )σx 2 (x, ξ∗ ) w −α (ξ∗ )h(x, ξ∗ ) dξ∗ dξ
R3
2 k(ξ, ξ∗ )ec ||ξ˜ |2 −|ξ∗ |2 | w −α (ξ∗ )h(x, ξ∗ ) dξ∗ dξ 3
sup
x∈ R
R3
k(ξ, ξ∗ )e 2c ||ξ˜ |2 −|ξ∗ |2 | w −2α (ξ∗ ) dξ∗ sup
c
x∈ R 3
R3
k(ξ, ξ∗ )h2 (x, ξ∗ ) dξ∗ dξ
R3
. c h L 2 ( L ∞ x )
(4.6)
ξ
1
Furthermore, with respect to the other part of
−1
σx2 K¯ σx 2 w −α , we have
1 12 σ + σx e P ξ1 σx− 2 e − σ w −α h22
Lξ (L∞ x )
0
2 1 − 12 + − σ sup σx2 (x, ξ )e σ (x, ξ )ξ∗1 χ + (ξ ) χ (ξ ) e ( x , ξ ) σ ( x , ξ ) w (ξ ) h ( x , ξ ) d ξ ∗ ∗ x ∗ −α ∗ ∗ ∗ dξ j j 3
=
x∈ R
c
j
R3
2 −c |ξ˜ |2 −c |ξ |2 ∗ sup e w −α (ξ∗ )h(x, ξ∗ ) dξ∗ dξ 3
x∈ R
R3
˜2 2 c e −2c |ξ | −2c |ξ∗ | w −2α (ξ∗ ) dξ∗ sup h2 (x, ξ∗ ) dξ∗ dξ 3 x∈ R
R3
. c h L 2 ( L ∞ x )
(4.7)
ξ
Then the proof is completed.
R3
2
734
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
Lemma 4.2. Review the properties of the operator U in [5] as follows, for sufficiently small and sufficiently large l, 1. U (h)(x, ξ ) L p c ν −1 h(x, ξ ) L p , x
x
2. U (h)(x, ξ ) L r ( L p ) c ν −1 h(x, ξ ) L r ( L p ) , ξ
x
3. U (h)(x, ξ ) σ β c ν −1 h(x, ξ ) σ β , 1
− 12
1 2
− 12
4. σx2 U σx 5. σx U σx
ξ
x
(h)(x, ξ ) L xp c ν −1 h(x, ξ ) L xp , (h)(x, ξ ) Lr (L xp ) c ν −1 h(x, ξ ) Lr (L xp ) , ξ
ξ
where 1 p , r ∞ and β ∈ R. Next, we will estimate Eq. (3.6) in order to obtain the estimate on · σ β of g. Lemma 4.3. The solution to the linearized Boltzmann equation with damping (3.6) satisfies for 0 α < 12 , 1 12 1 σx w −α ν 12 g + σx2 w 1 ν − 12 g x c |ξ1 | 12 w −α a0 + h¯ 2 + w −α σx2 h¯ . +
(4.8)
σx , we have 12 2 1 2 0 0 σx g 0 + ν 2 g 1 + ξ1 g , g c ( , a0 ) ξ1 a0 , a0 + + h¯ 2 . 2 −
Proof. Since Theorem 3.3 and the properties of
(4.9)
Let θ be a cut-off function for large |ξ |. Multiplying θ w −α σx g to (3.6) and integrating it with respect to x and ξ over R + × R 3 , we have 2
2
1
θ 2 w 2−α σx |ξ1 | g 0 , g 0 − + θ 2 w 2−α σx ν g , g 2 1 = θ 2 w 2−α σx2 |ξ1 | g , g + θ 2 w 2−α σxx |ξ1 | g , g + θ 2 w 2−α σx g , K¯ g 2 1
+ θ 2 w 2−α σx g , h¯ + θ 2 w 2−α σx |ξ1 | g 0 , g 0 + .
(4.10)
2
In the following, we only analyze some terms on the right side of the above equation since others are simple. Since 1
− 12
w −α σx2 K¯ σx
is a bounded operator from L 2ξ to L 2ξ , we have
1 1 −1 θ 2 w 2−α σx g , K¯ g θ 2 w −α σx2 g w −α σx2 K¯ σx 2 g 1 2 1 2 μθ σx2 w −α g + c σx2 g 1 2 μθ σx2 w −α g + c ξ1 a0 , a0 + + h¯ 22
(4.11)
where μ is a small constant. Since |σxx ξ1 | δ σx ν , we have
1 2
θ 2 w 2−α σxx |ξ1 | g , g δ θ 2 w 2−α νσx g , g .
(4.12)
What follows, we will estimate the last term of the right side of the above equality (4.10). When small, we have
1 2
θ 2 w 2−α σx ξ1 g 0 , g 0
= +
1
2 ξ 1 >0
c
2 θ 2 w 2−α σx ξ1 g 0 dξ
2 12 − σ (0,ξ ) −1 ξ M ∞ e θ 2 w 2−α σx ξ1 |a0 | + sM ∞2 M ω e σ (0,ξ ) g 0, ξ d ξ d ξ 1
ξ 1 >0
θ 2 w 2−α σx ξ1 |a0 |2 dξ + c
c
and s are sufficiently
ξ 1 >0
ξ1 <0
−1 2 2 σ (0,ξ ) ξ1 M ∞ ξ θ 2 w − 2 α σ x ξ1 s 2 M ∞ Mω e
ξ 1 >0 ξ <0
1 −1 2
w −2α ξ σx 0, ξ dξ dξ w −α σx ξ1 g 0 , g 0 − ×e
1 c θ 2 w 2−α σx ξ1 |a0 |2 dξ + w 2−α σx ξ1 g 0 , g 0 − .
−2 σ (0,ξ )
4
ξ 1 >0
(4.13)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
735
By letting θ → 1, the following inequality 1 12 σx w −α ν 12 g c |ξ1 | 12 w −α a0 + h¯ 2 + w −α σx2 h¯ +
(4.14)
holds. From (4.2) and the above inequality, we can get
2
2 1 1 12 σx w 1 ν − 12 g x 2 σx2 w 1 σx − ν (ξ ) g + σx2 w 1 1 (h¯ + K¯ g ) ξ1 ξ1 1 1 2 1 2 1 2 c σx2 ν 2 g + σx2 g + σx2 h¯ .
(4.15)
2
Combining (4.14) with (4.15) completes the proof.
After the above preparations, we can estimate the g σ β now. Lemma 4.4. For 0 < α < 12 , β > 32 , the solution to the problem (3.6) satisfies 1 1
g σ β c ν −1 h¯ σ β + h¯ 2 + w −α σx2 h¯ + a0 +,β + |ξ1 | 2 w −α a0 +
(4.16)
where a0 +,β = supξ ∈ R 3 ,ξ1 >0 (1 + |ξ |)β |a0 (ξ )|. Proof. From the formula g = a˜ + U ( K¯ g + h¯ ) and by using the properties of U and K¯ , we have
g σ β c ˜a σ β + ν −1 K¯ g σ β + ν −1 h¯ σ β c ˜a σ β + g σ (β−1) + ν −1 h¯ σ β .
(4.17)
Notice that
12 − σ (0,ξ ) − 12
σ (0,ξ ) exp −κ (x, ξ ) sM ∞ M e M e g 0 , ξ d ξ ξ ω ∞ 1 ξ1 <0
σβ
1 c |ξ1 | 2 g 0 − .
(4.18)
Then, inserting the above result into the estimate (4.17), the following
1
g σ β c a0 +,β + g σ (β−1) + ν −1 h¯ σ β + |ξ1 | 2 g 0 − 1 c a0 +,β + g σ 0 + ν −1 h¯ + |ξ1 | 2 g 0
(4.19)
−
σβ
holds. In order to estimate g σ 0 , by using the result of Lemma 4.3 and Theorem 3.3, we have for
12 σx w γ g 2∞
L x ( L 2ξ )
1 4
< γ < 12 ,
2 1 σx2 w γ g L 2 ( L ∞ ) ξ
x
1 1 1 1 1 c σx2 w −(1−2γ ) ν 2 g + σx2 w 1 ν − 2 g x + σx2 g 1 1 c |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ .
(4.20)
Using the expression (4.3), we have 1
1
1
− 12
σx2 g = σx2 a˜ + σx2 U σx
1
1
σx2 K¯ g + σx2 h¯ .
(4.21)
From Theorem 3.3, Lemma 4.1 and Lemma 4.2, we have
12 σ x g
L 2ξ ( L ∞ x )
1 1 c ˜a L 2 ( L ∞ + ν −1 σx2 K¯ g L 2 ( L ∞ ) + ν −1 σx2 h¯ L 2 ( L ∞ ) x )
ξ
ξ
x
ξ
x
1 1 1 1 −1 c a0 + + |ξ1 | 2 g 0 − + σx2 K¯ σx 2 w −γ σx2 w γ g L 2 ( L ∞ ) + ν −1 σx2 h¯ L 2 ( L ∞ )
ξ
x
ξ
x
1 1 1 c a0 + + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ + ν −1 σx2 h¯ L 2 ( L ∞ ) .
ξ
x
(4.22)
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J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
Similarly, we have 1 1 1
g σ 0 c a0 +,0 + |ξ1 | 2 g 0 − + ν −1 σx2 K¯ g L ∞ + ν −1 σx2 h¯ L ∞ x,ξ
x,ξ
1 1 1 c a0 +,0 + |ξ1 | 2 g 0 − + σx2 g L 2 ( L ∞ ) + ν −1 σx2 h¯ L ∞
ξ
x
x,ξ
1 1 c a0 + + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ 1 1 + ν −1 σx2 h¯ 2 ∞ + ν −1 σx2 h¯ .
(4.23)
σ0
Lξ (L x )
Finally, when β > 32 , we have 1 1
g σ ,β c a0 +,β + a0 + + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ 1 1 + ν −1 σx2 h¯ L 2 ( L ∞ ) + ν −1 σx2 h¯ σ 0 + ν −1 h¯ σ β
ξ
c a0 +,β This completes the proof.
x
1 1 + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ + ν −1 h¯ σ β .
(4.24)
2
After the above estimates about the solution to the linearized Boltzmann equation with damping, we will prove the existence of a solution of the nonlinear Boltzmann equation with damping. Below, we will introduce the following two lemmas on the properties of Γ ( g , g ), see [5]. Lemma 4.5. The projection of Γ ( g , g ) on the null space of L vanishes and there exists a positive constant c, such that
−1 σ − σ ν e Γ e g , e − σ g
2
2 3−β1
g 2σ β σ β c exp − (δ x + l)
for any β >
1−β1 2
(4.25)
and sufficiently small constant > 0.
Lemma 4.6. When β >
5 2
and is sufficiently small, we have the results below:
− 12 σ − σ σ x e Γ e g , e − σ g c g 2σ β , σ − σ e Γ e g , e − σ g 2 c g 2σ β , 1 w −α σx2 e σ Γ e − σ g , e − σ g c g 2 , σβ
0<α<
1 2
(4.26)
.
In summary, we have the following existence theorem for the nonlinear problem with damping. −1
Theorem 4.7. When the boundary data F 0 satisfies that M ∞2 e σ (0,ξ ) ( F 0 − M ∞ + sM ω ξ <0 |ξ1 | M ∞ (0, ξ ) dξ ) is sufficiently small 1 in the norms · +,β , T ω < 2T ∞ , l is sufficiently large, , δ and s are sufficiently small and l−1 , then Eq. (3.5) has a unique solution F , such that e σ W 0−1 ( F − M ∞ ) is bounded in the norm · σ β , when β > 52 .
Proof. For any h which has · σ β < ∞, there exists a unique solution to the equation according to the theorem,
⎧ x > 0, ξ ∈ R 3 , ξ1 g x − σx ξ1 g − L g − D g = e σ Γ e − σ h, e − σ h , ⎪ ⎪ ⎨ 12 − σ (0,ξ ) −1 ξ M ∞ e g |x=0 = a0 (ξ ) + sM ∞2 M ω e σ (0,ξ ) g 0 , ξ d ξ , ξ1 > 0 . 1 ⎪ ⎪ ⎩
(4.27)
ξ 1 <0
Furthermore, from Lemmas 4.4 and 4.6, we will get
1
g σ β c h 2σ β + a0 +,β + |ξ1 | 2 w −α a0 + c h 2σ β + a0 +,β
(4.28)
since β > 52 . Hence, when a0 +,β is sufficiently small, there exists a solution g to (3.5) by contraction mapping theorem. Since g satisfies the inequality
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
737
Table 1
M∞ < − 1
− 1 < M∞ < 0 0 < M∞ < 1
M∞
>1
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 0
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 1
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 4
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 5
g σ β c g 2σ β + a0 +,β , the solution is unique.
2
As in [16], we will give an implicit solvability condition on the boundary data for the existence of a solution to the nonlinear Boltzmann equation. Let A γ be a nonlinear operator
A γ (b0 ) ≡ f (0, ·)
(4.29)
where f (x, ξ ) = e − σ (x,ξ ) g is given by
⎧ x > 0, ξ ∈ R 3 , ξ1 f x − L f = h − γ (δ x + l)−Θ P 0+ ξ1 f , ⎪ ⎪ ⎨ 12 −1 ξ M ∞ f 0, ξ dξ , ξ1 > 0. f |x=0 = b0 (ξ ) + sM ∞2 M ω 1 ⎪ ⎪ ⎩
(4.30)
ξ 1 <0
From the energy estimate on the solution to the nonlinear Boltzmann equation with damping, we know the operator satisfies
A γ : e σ (0,ξ ) b0 ∈ L 2ξ1 ,+ → A γ (b0 ) ∈ L 2|ξ | , and is bounded when b0 is sufficiently small. It is easy to see that the solution of (3.4) is a solution of (1.1) if P 0+ ξ1 A γ (b0 ) ≡ 0. Using the method in [16], we can get the key result in this paper. Theorem 4.8. The classification of the boundary data near the far field Maxwellian satisfying the solvability condition P 0+ ξ1 A γ (b0 ) ≡ 0 with respect to Mach number in the far field can be summarized in Table 1. References [1] K. Aoki, Y. Sone, T. Yamada, Numerical analysis of gas flows condensing on its plane condensed phase on the basis of kinetic theory, Phys. Fluids A 2 (1990) 1867–1878. [2] K. Aoki, K. Nishino, Y. Sone, H. Sugimoto, Numerical analysis of steady flows of a gas condensing on or evaporating from its plane condensed phase on the basis of kinetic theory: Effect of gas motion along the condensed phase, Phys. Fluids A 3 (1991) 2260–2275. [3] C. Bardos, R.E. Caflish, B. Nicolaenko, The Milne and Kramers problems for the Boltzmann equation of a hard sphere gas, Comm. Pure Appl. Math. 39 (1986) 323–352. [4] C. Cercignani, R. Illner, M. Pulvirenti, The Mathematical Theory of Dilute Gases, Appl. Math. Sci., Springer-Verlag, New York, 1994. [5] C.C. Chen, T.P. Liu, T. Yang, Exsistence of boundary layer solutions to the Boltzmann equation, Anal. Appl. 2 (4) (2004) 337–363. [6] F. Coron, F. Golse, C. Sulem, A classification of well-posed kinetic layer problems, Comm. Pure Appl. Math. 41 (1988) 409–435. [7] R.T. Glassey, The Cauchy Problem in Kinetic Theory, Society for Industrial and Applied Mathematics (SIAM), Philadephia, 1996. [8] F. Golse, B. Perthame, C. Sulem, On a boundary layer problem for the nonlinear Boltzmann equation, Arch. Ration. Mech. Anal. 103 (1) (1988) 81–96. [9] F. Golse, F. Poupaud, Stationary solutions of the linearized Boltzmann equation in a half-space, Math. Methods Appl. Sci. 11 (1989) 483–502. [10] H. Grad, Asymptotic theory of the Boltzmann equation, in: J.A. Laurmann (Ed.), Rarefied Gas Dynamics, vol. 1, Academic Press, New York, 1963, pp. 26– 59. [11] Y. Sone, Kinetic Theory and Fluid Dynamics, Birkhäuser, Berlin, 2002. [12] Y. Sone, Kinetic theory of evaporation and condensation linear and nonlinear problems, J. Phys. Soc. Japan 45 (1) (July 1978). [13] Q.Z. Tian, Existence of nonlinear boundary layer solution to the Boltzmann equation with physical boundary conditions, J. Math. Anal. Appl. 356 (1) (2009) 42–59. [14] Q.Z. Tian, J. Sun, Nonlinear stability of boundary layer solution to the Boltzmann equation with diffusive effect at the boundary, J. Math. Phys. 50 (2009) 103303 (SCI). [15] S. Ukai, K. Asano, On the Cauchy problem of the Boltzmann equation with a soft potential, Publ. Res. Inst. Math. Sci. 18 (2) (1982) 477–519. [16] S. Ukai, T. Yang, S.H. Yu, Nonlinear boundary layers of the Boltzmann equation: I. Existence, Comm. Math. Phys. 236 (3) (2003) 373–393. [17] S. Ukai, T. Yang, S.H. Yu, Nonlinear stability of boundary layers of the Boltzmann equation. I. The case M∞ < −1, Comm. Math. Phys. 244 (1) (2004) 99–109. [18] W.K. Wang, T. Yang, X.F. Yang, Existence of boundary layer to the Boltzmann equation with cutoff soft potentials, J. Math. Phys. 48 (7) (2007). [19] W.K. Wang, T. Yang, X.F. Yang, Nonlinear stability of boundary layers of the Boltzmann equation for cutoff hard potentials, J. Math. Phys. 47 (8) (2006). [20] X.F. Yang, Nonlinear stability of boundary layers for the Boltzmann equation with cutoff soft potentials, J. Math. Anal. Appl. 345 (2) (2008) 941–953.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
The nonlinear boundary layer to the Boltzmann equation with mixed boundary conditions for hard potentials Jie Sun a,b,c,∗ , Qianzhu Tian a,b,c,1 a b c
Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China Department of Mathematics, City University of Hong Kong, Kowloon, Hong Kong Joint Advanced Research Center of University of Science and Technology of China and City University of Hong Kong, Suzhou, Jiangsu 215123, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 18 April 2010 Available online 13 October 2010 Submitted by Steven G. Krantz
In this paper, the existence of boundary layer solutions to the Boltzmann equation for hard potential with mixed boundary condition, i.e., a linear combination of Dirichlet boundary condition and diffuse reflection boundary condition at the wall, is considered. The boundary condition is imposed on the incoming particles, and the solution is supposed to approach to a global Maxwellian in the far field. As for the problem with Dirichlet boundary condition (Chen et al., 2004 [5]), the existence of a solution highly depends on the Mach number of the far field Maxwellian. Furthermore, an implicit solvability condition on the boundary data which shows the codimension of the boundary data is related to the number of the positive characteristic speeds is also given. © 2010 Elsevier Inc. All rights reserved.
Keywords: Boltzmann equation Boundary layer Hard potential Mixed boundary condition
1. Introduction In our paper, we consider the boundary layer problem of the Boltzmann equation for hard potential with angular cutoff which arises in the physical condensation–evaporation problem. We will study the stationary problem in a half-space with a mixed boundary condition, i.e., a linear combination of Dirichlet boundary condition and diffuse reflection boundary condition and convergence to an equilibrium state at infinity. A lot of numerical results about the boundary layer problem have been made by [1,2,11,12]. The boundary layer problem for linearized Boltzmann equation has been well-studied in [3,6,9]. The existence of solutions was solved in [8] on the nonlinear problem with specular reflection boundary condition. With respect to the nonlinear problem with Dirichlet boundary conditions, existence and stability of the boundary layer solutions are given by [5,16,18] and [17,19,20] respectively. Recently, the problems with mixed boundary conditions for hard sphere case were considered in [13,14]. In our paper, we will work on the boundary layer problem with a mixed boundary condition by using the method developed in [5] for hard potential case. The boundary layer problem we consider is as follows,
⎧ x > 0, ξ ∈ R 3 , ξ1 F x = Q ( F , F ), ⎪ ⎪ ⎪ ⎨ ξ F 0 , ξ d ξ , ξ1 > 0 , F |x=0 = F 0 (ξ ) + sM ω 1 ⎪ ⎪ ξ1 <0 ⎪ ⎩ F → M ∞ (x → ∞), ξ ∈ R3.
*
(1.1)
Corresponding author at: Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China. E-mail addresses: [email protected] (J. Sun), [email protected] (Q. Tian). 1 The author’s research was supported both by the National Natural Science Foundation of China Grant No. 10871187, and the Natural Science Foundation of Anhui Grant No. 070416228. 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2010.09.067
©
2010 Elsevier Inc. All rights reserved.
726
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
As usual, the collision operator takes the form
F ξ F ξ∗ − F (ξ ) F (ξ∗ ) q( v , θ) dξ∗ dω,
Q (F , F ) = R3×S2
where ξ and ξ∗ are velocities of gas particles before collision, while ξ = ξ − ω(ξ − ξ∗ ) · ω and ξ∗ = ξ∗ + ω(ξ − ξ∗ ) · ω are velocities after collision. Here v = ξ − ξ∗ , ω ∈ S 2 and cos θ = v| v·ω| . This is a stationary problem with mixed boundary condition at x = 0 for incoming particles, where M ∞ (ξ ) = M [ρ∞ ,u ∞ , T ∞ ] (ξ ) and M ω (ξ ) = M [1,0, T ω ] (ξ ) are Maxwellians. As to the collision kernel q( v , θ), we have the following assumption for the hard potential with angular cutoff, cf. [15]. Assumption. There exist 0 δ < 1 and a positive constant c such that
0 q( v , θ) c | v | + | v |−δ | cos θ|;
(1.2)
there is a positive constant c,
e−
|ξ∗ |2 2
q( v , θ) dξ∗ dω
R3×S2
e −|ξ∗ | q( v , θ) dξ∗ dω 2
−1
c.
R3×S2
In this paper, the gas constant R is normalized to 1. Then a Maxwellian with the mass density temperature T is given by
M=
(1.3)
ρ (2π T )
3 2
e−
|ξ −u |2 2π T
ρ , flow velocity u and
.
It is well known that Q ( M ) = 0. Furthermore, the collision operator has exactly five collision invariants
φ 0 = 1,
φi = ξ
φ4 = |ξ |2
(i = 1, 2, 3),
satisfying
φ i , Q ( F ) = 0,
i = 0, 1 , . . . , 4 ,
for any distribution function F (see [7]), here , is an inner product of L 2ξ . Without loss of generality, we can assume u ∞,2 and u ∞,3 of the Maxwellian in the far field are zeroes. Then the sound speed and Mach number of the far field Maxwellian can be defined
c∞ =
5 3
M∞ =
T ∞,
u ∞,1 c∞
(1.4)
, 1
2 see [4]. We will consider the boundary layers near the global Maxwellian M ∞ at infinity. Let F = M ∞ + M ∞ f , then Eq. (1.1) will be reformulated into
⎧ x > 0, ξ ∈ R 3 , ⎪ ⎪ ξ1 f x − L f = Γ ( f , f ), ⎪ ⎪ 1 1 1 ⎨ − − 2 ξ M ∞ + M ∞ f |x=0 = M ∞2 ( F 0 − M ∞ ) + sM ∞2 M ω f 0 , ξ d ξ , ξ1 > 0 , 1 ⎪ ⎪ ξ1 <0 ⎪ ⎪ ⎩ f → 0 (x → ∞), ξ ∈ R 3,
(1.5)
where 1 − 1 12 2 Γ ( f , f ) = M ∞2 Q M ∞ f , M∞ f
and −1
1
1
2 2 L f = M ∞2 Q M ∞ , M ∞ f + Q M∞ f , M∞ .
In the following, we will introduce two weight functions. The first one is
−β
W β (ξ ) = 1 + |ξ |
M 1/2 [1, u ∞ , T ∞ ](ξ )
(1.6)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
with β ∈ R. The second one is
δx + l (1 + |ξ − u ∞ |)3−β1
δx + l δx + l 2 + + 3|ξ − u ∞ | η (1 + |ξ − u ∞ |)1−β1 (1 + |ξ − u ∞ |)3−β1 2
727
σ (x, ξ ) = 5(δ x + l) 3−β1 1 − η
where x 0 and
η (s) =
(1.7)
η : (0, ∞) → R is a smooth non-increasing function which has the following properties: s 1, s 2,
1, 0,
and 0 η(s) 1. After the introduction of the above notations, we state our main result in this paper. Theorem 1.1. Suppose M∞ = 0, ±1 and β > 5/2. If s is small enough and T ω < 2T ∞ , then there exist positive numbers , and a C 1 map
+ Ψ : L 2 R 3+ , e σ (0,ξ ) ξ1 dξ → R n ,
Ψ (0) = 0,
1 , 2 , (1.8)
such that the following holds. (i) For any F 0 satisfying
σ (0,ξ ) 1/2 1 W β , e F − M + sM M 0 , ξ d ξ ξ 0 ∞ ω ∞ 1
ξ ∈ R3+ ,
(1.9)
ξ <0 1
Eq. (1.1) admits a unique solution F in the class
σ (x,ξ ) −1/2 e M ∞ ( F − M ∞ )σ β 2 ,
(1.10)
when F 0 satisfies
1/2 −1/2 ξ M ∞ 0, ξ dξ =0 Ψ M∞ F 0 − M ∞ + sM ω 1
(1.11)
ξ <0 1
where l > 0 is large and δ > 0 is small in σ (x, ξ ) and · σ β is defined in (4.1). (ii) The set of F 0 satisfying (1.9) and (1.11) forms a local C 1 manifold of co-dimension n+ . Remark 1.2. The number n+ of solvability conditions changes with Mach number as
⎧ 0, ⎪ ⎪ ⎪ ⎨ 1, n+ = ⎪ 4, ⎪ ⎪ ⎩ 5,
M∞ ∈ (−∞, −1), M∞ ∈ (−1, 0),
(1.12)
M∞ ∈ (0, 1), M∞ ∈ (0, ∞).
We exclude the cases M∞ = 0, ±1 which are also important physically. 2. Preparations Before we study the equation itself, we will study the properties of the weight functions and the linearized collision operator. 2.1. Weight functions
σ ⎧ c 1 (1 + |ξ − u ∞ |)2 , ⎪ ⎪ ⎨ 2 σ (x, ξ ) = c2 (δ x + l) 3−β1 + c3 (1 + |ξ − u ∞ |)2 , ⎪ ⎪ 2 ⎩ 5(δ x + l) 3−β1 ,
A careful analysis helps us to get the properties of
below:
(x, ξ ) ∈ Ω1 , (x, ξ ) ∈ Ω2 , (x, ξ ) ∈ Ω3 ,
(2.1)
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where
⎧ Ω1 = (x, ξ ): ⎪ ⎪ ⎨ Ω2 = (x, ξ ): ⎪ ⎪ ⎩ Ω3 = (x, ξ ):
3−β1 , δ x + l 1 + |ξ − u ∞ | 3−β1 3−β1 , 1 + |ξ − u ∞ | δ x + l 2 1 + |ξ − u ∞ | 3−β1 δ x + l 2 1 + |ξ − u ∞ | ,
c 1 , c 2 and c 3 are positive functions of (x, ξ ), and c 1 and c 2 + c 3 have uniform lower bounds. With respect to σx , we have the following results:
⎧ δ(1 + |ξ − u ∞ |)−1+β1 , (x, ξ ) ∈ Ω1 , ⎪ ⎪ ⎪ ⎨ 1−β − 3−β1 − 1+β1 1 ), + c 5 (δ x + l) (x, ξ ) ∈ Ω2 , σx = δ(c4 (1 + |ξ − u ∞ |) ⎪ ⎪ 1−β1 ⎪ ⎩ 10δ (δ x + l)− 3−β1 , (x, ξ ) ∈ Ω3 , 3−β1
(2.2)
where c 4 + c 5 has a uniform lower bound. From the above results, we can learn 2
σ (x, ξ ) c6 (δ x + l) 3−β1 ,
1−β
− 3−β1
0 < c 7 min (δ x + l)
1
1−β1 −1+β1 − σx c 8 (δ x + l) 3−β1 , , 1 + |ξ − u ∞ |
β |σx ξ1 | c 1 + |ξ | 1 c ν ,
(2.3)
where c 6 , c 7 and c 8 are positive constants. 2.2. Linearized collision operator It is well known that the linearized collision operator L is of the form:
L f = − f (ξ )
1
2 M ∞ (ξ∗ )q( v , θ) dξ∗ dω − M ∞
1 2 f (ξ∗ ) M ∞ (ξ∗ )q( v , θ) dξ∗ dω
1 12 12 2 + f ξ∗ M ∞ ξ + f ξ M ∞ ξ∗ M ∞ (ξ∗ )q( v , θ) dξ∗ dω = −ν (ξ ) − K 1 + K 2 f (ξ ).
(2.4)
In the sequel, we assume that
β 1
c 1 1 + |ξ |
β 2 ν (ξ ) c 2 1 + |ξ |
for constants 0 < β1 β2 1. K 1 and K 2 are integral compact operators with kernels k1 (ξ, ξ∗ ) and k2 (ξ, ξ∗ ) respectively. Set K = − K 1 + K 2 and k = −k1 + k2 , cf. [10]. The kernel k1 has the following property:
k1 (ξ, ξ∗ ) |ξ − ξ∗ | + |ξ − ξ∗ |
−δ0
|ξ − u ∞ |2 |ξ∗ − u ∞ |2 exp − − 4T ∞
4T ∞
where 0 δ0 < 1. Furthermore, the kernel k2 (ξ, ξ∗ ) satisfies,
k2 (ξ, ξ∗ ) = a(ξ, ξ∗ ) exp −
||ξ − u ∞ |2 − |ξ∗ − u ∞ ||2 |ξ∗ − ξ |2 − 8T ∞ 8T ∞ |ξ∗ − ξ |2
(2.5)
(2.6)
where
a(ξ, ξ∗ ) c |ξ − ξ∗ |−1 .
(2.7)
Then, k has the following properties:
R3
k(ξ, ξ∗ ) 1 + |ξ∗ | −β dξ∗ 1 + |ξ | −β−1 ,
β ∈ R,
k(ξ, ξ∗ )2 dξ c ,
R3
∞ ∞ −∞ −∞
k(ξ, ξ∗ ) dξ2 dξ3 c .
(2.8)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
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At the end of this section, we will state two lemmas which have been proved in [5]. Lemma 2.1. There is a constant 1 > 0, such that for 0 1 and g ∈ N ⊥ ,
2 2 − g , e |ξ −u ∞ | Le − |ξ −u ∞ | g ν1 ν |ξ | g , g ,
(2.9)
for a positive constant ν1 depending on 1 . Lemma 2.2. There is a constant 2 > 0, such that for 0 2 and g ∈ N ⊥ ,
− g , e σ (x,ξ ) Le − σ (x,ξ ) g ν2 ν |ξ | g , g ,
(2.10)
for a positive constant ν2 depending on 2 . 3. Linear existence In order to solve Eq. (1.1), we will firstly show the existence of a solution to the linearized problem (3.5) with damping term. Define the following orthogonal projections with respect to L:
P 0 : L 2ξ → N ,
P 1 = I − P 0 : L 2ξ → N ⊥
(3.1) 1
2 where N is the null space of the linearized collision operator L and is spanned by the five collision invariants M ∞ φi , i = 0, 1, . . . , 4. N ⊥ denotes the orthogonal complement of N in L 2ξ . Define the operator for macroscopic convection:
A = P 0 ξ1 P 0 .
(3.2)
It is easy to see that A has the eigenvalues
λ1 = u ∞,1 − c ∞ ,
λi = u ∞,1 (i = 2, 3, 4),
λ5 = u ∞,1 + c ∞ ,
(3.3)
on N. n+ happens to be the number of the positive eigenvalues. As in [16], we decompose the operator A into the positive part and the negative part A + , A − , and denote the corresponding projections by P + , P − . Note that if M ∞ = 0, ±1, then 0
+
−
A=A +A ,
+
0
−
P0 = P0 + P0 .
We modify (1.1) and (1.5) by adding a damping term,
and
⎧ ξ1 F x = Q ( F , F ) − γ (δ x + l)−Θ P 0+ ξ1 W 0−1 ( F − M ∞ ), x > 0, ξ ∈ R 3 , ⎪ ⎪ ⎪ ⎪ ⎨ ξ F 0, ξ d ξ , F |x=0 = F 0 (ξ ) + sM ω ξ1 > 0 , 1 ⎪ ⎪ ξ 1 <0 ⎪ ⎪ ⎩ F → M ∞ (x → ∞), ξ ∈ R3,
(3.4)
⎧ x > 0, ξ ∈ R 3 , ξ1 f x − L f = h − γ (δ x + l)−Θ P 0+ ξ1 f , ⎪ ⎪ ⎪ ⎪ 12 ⎨ −1 ξ M ∞ f 0, ξ dξ , ξ1 > 0, f |x=0 = b0 (ξ ) + sM ∞2 M ω 1 ⎪ ⎪ ξ1 <0 ⎪ ⎪ ⎩ f → 0 (x → ∞), ξ ∈ R3,
(3.5)
where −1
−1
b0 (ξ ) = M ∞2 ( F 0 − M ∞ ) + sM ∞2 M ω h = Γ ( f , f ),
Θ=
1 − β1 3 − β1
ξ M ∞ dξ , 1
ξ1 <0
,
and γ is a positive constant to be chosen later. Note that for the case M ∞ < −1, A has no positive eigenvalue. Hence, P 0+ = 0 and the damping term vanishes. Let
f = exp − σ (x, ξ ) g .
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The problem (3.5) turns into
⎧ x > 0, ξ ∈ R 3 , ξ g − σx ξ1 g − L g + D g = h¯ , ⎪ ⎪ ⎨ 1 x 1 1 − 2 ξ M ∞ g |x=0 = a0 (ξ ) + sM ∞2 M ω e σ (0,ξ ) e − σ (0,ξ ) g 0, ξ dξ , ξ1 > 0, 1 ⎪ ⎪ ⎩
(3.6)
ξ 1 <0
where
−1
a0 (ξ ) = M ∞2 e σ (0,ξ ) F 0 − M ∞ + sM ω
ξ M ∞ 0, ξ dξ 1
ξ1 <0
and
h¯ = e σ (x,ξ ) h,
D = γ (δ x + l)−Θ e σ P 0+ ξ1 e − σ .
L = e σ Le − σ ,
In the following, we will mainly consider the linear problem (3.6). Let D ∗ be the adjoint operator of D , and set
ψ = −ξ1 φx − σx ξ1 φ − L − φ + D ∗ φ, φ ∈ C 0∞ , 1 1 V = ψ 0 − 2 φ = φ|x=0 = sM ∞ e − σ ξ >0 |ξ1 | M ∞2 M ω e σ φ 0 dξ for ξ1 < 0. 1
(3.7)
We define
g˜ ψ = (h¯ , φ) +
1 2
ξ1 a 0 , φ 0
+
where (,) denotes the inner product in L 2 ( R + × R 3 ) and f , g ± = ±ξ >0 f g dξ . 1 Consider
ψ(φ), φ =
1 − σx ξ1 φ 2 − φ L φ + φ D φ dξ dx +
2 ξ1 φ 0 d ξ
2
R+ R3
+ s2
ξ1 M ∞ e −2 σ
ξ 1 <0
ξ 1 >0
− 12 ξ M ∞ M ω e σ φ 0 dξ 1
2 dξ.
(3.8)
ξ1 >0
In order to estimate the above term, we show the following lemma first. Lemma 3.1. Let l be sufficiently large. Then there exists a positive constant c, such that when γ = O (1) 1 and γ > ,
|ξ˜ |
3−β1
φ γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ dξ c
φ 2 dξ,
(3.9)
R3
2l
for any φ ∈ N where ξ˜ = ξ − u ∞ . Proof. We set φ =
5
χ j , where χ j is the normalized eigenvector of A corresponding to the eigenvalue λ j . Then γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ = (γ − )λ+j b j χ + − λ−j b j χ − , j j j =1 b j
− where λ+ j (λ j ) denotes the positive (negative) eigenvalue, and Therefore, when l is sufficiently large,
− + − χ+ j (χ j ) denotes the corresponding eigenvector to λ j (λ j ).
5 φ γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ dξ < μ b2j = μ i =1
|ξ˜ |3−β1 2l
φ 2 dξ. R3
Moreover, since λ = 0 and
γ = O (1) , there exists a positive constant c, such that φ γ P 0+ ξ1 P 0 − P 0 ξ1 P 0 φ dξ c φ 2 dξ.
R3
When l is sufficiently large, we have
R3
μ < 2c . Then the proof is completed. 2
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
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By using Lemma 3.1, we can estimate the term (ψ(φ), φ) as follows. Lemma 3.2. Let l be sufficiently large, and δ be sufficiently small and l−1 . When s is sufficiently small and T ω < 2T ∞ , there exists a constant c, such that when γ = O (1) 1 and γ > , the following result holds
2 1 2
ψ(φ), φ c (δ x + l)−Θ/2 φ0 + ν 2 φ1 + ξ1 φ 0 , φ 0 + .
Proof. Let φ1 = P 1 φ, and φ0 =
5
φ D φ − ξ1 σx φ02 dξ =
R3
j =1 b j
R3
χ j . We consider the term
R3
(3.10)
(φ D φ − ξ1 σx φ02 ) dξ at first,
φ γ (δ x + l)−Θ P 0+ ξ1 P 0 φ − ξ1 σx φ02 dξ
+
φ D − γ (δ x + l)−Θ P 0+ ξ1 P 0 φ dξ.
(3.11)
R3
Notice that
ξ1 σx φ02 dξ R3
10δ 3 − β1
(δ x + l)−Θ
5 2 φ0 P 0 ξ1 φ0 dξ + c (δ x + l)−Θ exp −c (δ x + l) 3−β1 b2j , j =1
2(1+|ξ˜ |)3−β1 δ x+l
then from Lemma 3.1, we have
φ γ (δ x + l)−Θ P 0+ ξ1 P 0 φ − ξ1 σx φ02 dξ c (δ x + l)−Θ φ0 2L 2 .
(3.12)
ξ
R3
We know that
φ D − γ (δ x + l)−Θ P 0+ ξ1 P 0 φ dξ = γ (δ x + l)−Θ φ, e σ P 0+ ξ1 e − σ − P 0+ ξ1 φ + φ, P 0+ ξ1 φ1 .
(3.13)
R3
As the kernel of (e σ P 0+ ξ1 e − σ − P 0+ ξ1 ) is
χ+ (ξ∗ )χ + (ξ )ξ∗1 e (σ (ξ )−σ (ξ∗ )) − 1 , j j
(3.14)
j
we can get that
c φ D − γ (δ x + l)−Θ P 0+ ξ1 P 0 φ dξ (δ x + l)−Θ φ0 2L 2 + c (δ x + l)−Θ φ1 2L 2 . 2
ξ
ξ
R3
From (2.3), we have the following result
1 2 c φ D φ − ξ1 σx φ 2 dξ (δ x + l)−Θ φ0 2L 2 − c (δ x + l)−Θ ν 2 φ1 L 2 . 2
ξ
ξ
R3
Secondly, similar to the case with Dirichlet boundary condition, we also have
1 2 φ L φ dξ −c 2 (δ x + l)−θ φ0 2L 2 + c ν 2 φ1 L 2 .
−
(3.15)
ξ
ξ
R3
Furthermore, when s is sufficiently small, we can have
1 2
s2 ξ 1 <0
ξ1 M ∞ e −2 σ
0 − 12 ξ φ M ∞ M ω e σ dξ 1
ξ1 >0
2 dξ
1 4
ξ1 φ 0 , φ 0 + .
(3.16)
In conclusion, we can get the following result
2 1 2
ψ(φ), φ c (δ x + l)−Θ/2 φ0 + ν 2 φ1 + ξ1 φ 0 , φ 0 +
for some positive constant c. The proof is ended.
2
(3.17)
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In order to achieve existence of a solution to the linear problem (3.6), we define the following norms:
2 1 Θ
u 1 = (δ x + l)− 2 u 0 + u 1 2 2 , 2 1 Θ
u 2 = (δ x + l) 2 u 0 + u 1 2 2 ,
(3.18)
where u 0 = P 0 u and u 1 = P 1 u. Thus, Lemma 3.2 implies that
1 ψ(φ) c φ 1 + c ξ1 φ 0 , φ 0 2 . 2 +
(3.19)
In the following, we get the key result for this section. Theorem 3.3. If h¯ 2 < ∞, T ω < 2T ∞ , , δ and s are sufficiently small, l is sufficiently large and l−1 , then the linearized problem with damping (3.6) has a unique solution g which satisfies
(δ x + l)− Θ2 g 0 2 + ν 12 g 1 2 + ξ1 g 0 , g 0 c ( , a0 ) ξ1 a0 , a0 + + h¯ 2 . 2 −
(3.20)
Proof. By (3.17), we have
1 | g˜ ψ| = (h¯ , φ) + ξ1 a0 , φ 0 + 2
1 0 ¯ ¯ = (h0 , φ0 ) + (h1 , φ1 ) + ξ1 a0 , φ + 2
1 c ( , a0 ) |ξ1 | 2 a0 + + h¯ 2 ψ 2 .
(3.21)
Then, g˜ can be taken as a continuous functional on the subspace V with respect to the norm · 2 . Hence, from the Hahn– Banach theorem and the Riesz representation theorem, we know there exists an element g in the space L 2 ( R + × R 3 ) such that ( g , ψ) = g˜ ψ . A similar argument to [16] yields that g is the unique solution to Eq. (3.6). Multiplying (3.6) by g and integrating the result over R + × R 3 , similar to the discussion in Lemma 3.2, we can get the estimate (3.20). 2 4. Nonlinear existence In this section, we will prove the existence for the nonlinear problem with damping on the basis of the estimates on the solution to the linearized equation (3.6). We will consider the problem in another space. Define the weighted norm:
1 β
f σ β = σx2 1 + |ξ | f L ∞ = x,ξ
1
sup x∈ R ,ξ ∈ R 3
β σx2 1 + |ξ | f (x, ξ ).
(4.1)
In order to obtain the above · σ β norm of the solution to Eq. (3.6), we will introduce the weighted function w α . Set
wα =
|ξ1 |α , |ξ1 | < 1, |ξ1 | 1.
1,
Rewrite (3.6)
g x = σx g +
1
ξ1
L g +
1 ¯ D h− g=
ξ1
ξ1
σx −
ν (ξ ) ξ1
g+
1 ¯ (h + K¯ g )
ξ1
(4.2)
where
K¯ = e σ K e − σ − γ e σ P 0+ ξ1 e − σ . Let
κ (x, ξ ) =
x
ν (ξ ) − σx + dy . ξ1 0
Therefore, the solution to Eq. (3.6) can be formally written as
g = a˜ + U ( K¯ g + h¯ ),
(4.3)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
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where
a˜ =
− 2 exp(−κ (x, ξ ))(a0 (ξ ) + sM ∞2 M ω e σ (0,ξ ) ξ <0 |ξ1 | M ∞ e − σ (0,ξ ) g (0, ξ ) dξ ), 1
ξ1 > 0 ,
0,
ξ1 < 0
1
and
x 0
U (h) =
−
1
exp(−(κ (x, ξ ) − κ (τ , ξ ))) ξ1 h(τ , ξ ) dτ ,
ξ1 > 0 ,
1
∞
1
exp(−(κ (x, ξ ) − κ (τ , ξ ))) ξ h(τ , ξ ) dτ ,
x
1
(4.4)
(4.5)
ξ1 < 0 .
The operators K¯ and U have the following properties stated in Lemma 4.1 and Lemma 4.2. Lemma 4.1. When is sufficiently small and l is sufficiently large, K¯ satisfies 1. 2.
1
− 12
1 2
− 12
σx2 K¯ σx σx K¯ σx
is a bounded operator from L 2ξ to itself. is a bounded operator from L 2ξ to L ∞ ξ .
3. K¯ h σ β c h σ (β−1) for β ∈ R . 1
− 12
4. w −α σx2 K¯ σx 1 2
− 12
1
−1
is a bounded operator from L 2ξ to itself, when 0 α < 12 .
5.
1 ∞ σx K¯ σx w −α is a bounded operator from L 2ξ ( L ∞ x ) to L x,ξ , when 0 α < 2 .
6.
1 σx2 K¯ σx 2 w −α is a bounded operator from L 2ξ ( L ∞ x ) to itself, when 0 α < 2 .
Proof. The details of the first five properties can be found in [5]. We will prove the sixth property in detail. 1
As
−1
σx2 (x, ξ )e σ (x, ξ )e − σ (x, ξ∗ )σx 2 (x, ξ∗ ) can be controlled by ec ||ξ˜ |
2
−|ξ∗ |2 | , for any h
∈ L 2ξ ( L ∞ x ), we have
1 12 σ σx e K σx− 2 e − σ w −α h22
Lξ (L∞ x )
1
sup
=
x∈ R 3
c
−1
2
σx2 (x, ξ )e σ (x, ξ )k(ξ, ξ∗ )e− σ (x, ξ∗ )σx 2 (x, ξ∗ ) w −α (ξ∗ )h(x, ξ∗ ) dξ∗ dξ
R3
2 k(ξ, ξ∗ )ec ||ξ˜ |2 −|ξ∗ |2 | w −α (ξ∗ )h(x, ξ∗ ) dξ∗ dξ 3
sup
x∈ R
R3
k(ξ, ξ∗ )e 2c ||ξ˜ |2 −|ξ∗ |2 | w −2α (ξ∗ ) dξ∗ sup
c
x∈ R 3
R3
k(ξ, ξ∗ )h2 (x, ξ∗ ) dξ∗ dξ
R3
. c h L 2 ( L ∞ x )
(4.6)
ξ
1
Furthermore, with respect to the other part of
−1
σx2 K¯ σx 2 w −α , we have
1 12 σ + σx e P ξ1 σx− 2 e − σ w −α h22
Lξ (L∞ x )
0
2 1 − 12 + − σ sup σx2 (x, ξ )e σ (x, ξ )ξ∗1 χ + (ξ ) χ (ξ ) e ( x , ξ ) σ ( x , ξ ) w (ξ ) h ( x , ξ ) d ξ ∗ ∗ x ∗ −α ∗ ∗ ∗ dξ j j 3
=
x∈ R
c
j
R3
2 −c |ξ˜ |2 −c |ξ |2 ∗ sup e w −α (ξ∗ )h(x, ξ∗ ) dξ∗ dξ 3
x∈ R
R3
˜2 2 c e −2c |ξ | −2c |ξ∗ | w −2α (ξ∗ ) dξ∗ sup h2 (x, ξ∗ ) dξ∗ dξ 3 x∈ R
R3
. c h L 2 ( L ∞ x )
(4.7)
ξ
Then the proof is completed.
R3
2
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Lemma 4.2. Review the properties of the operator U in [5] as follows, for sufficiently small and sufficiently large l, 1. U (h)(x, ξ ) L p c ν −1 h(x, ξ ) L p , x
x
2. U (h)(x, ξ ) L r ( L p ) c ν −1 h(x, ξ ) L r ( L p ) , ξ
x
3. U (h)(x, ξ ) σ β c ν −1 h(x, ξ ) σ β , 1
− 12
1 2
− 12
4. σx2 U σx 5. σx U σx
ξ
x
(h)(x, ξ ) L xp c ν −1 h(x, ξ ) L xp , (h)(x, ξ ) Lr (L xp ) c ν −1 h(x, ξ ) Lr (L xp ) , ξ
ξ
where 1 p , r ∞ and β ∈ R. Next, we will estimate Eq. (3.6) in order to obtain the estimate on · σ β of g. Lemma 4.3. The solution to the linearized Boltzmann equation with damping (3.6) satisfies for 0 α < 12 , 1 12 1 σx w −α ν 12 g + σx2 w 1 ν − 12 g x c |ξ1 | 12 w −α a0 + h¯ 2 + w −α σx2 h¯ . +
(4.8)
σx , we have 12 2 1 2 0 0 σx g 0 + ν 2 g 1 + ξ1 g , g c ( , a0 ) ξ1 a0 , a0 + + h¯ 2 . 2 −
Proof. Since Theorem 3.3 and the properties of
(4.9)
Let θ be a cut-off function for large |ξ |. Multiplying θ w −α σx g to (3.6) and integrating it with respect to x and ξ over R + × R 3 , we have 2
2
1
θ 2 w 2−α σx |ξ1 | g 0 , g 0 − + θ 2 w 2−α σx ν g , g 2 1 = θ 2 w 2−α σx2 |ξ1 | g , g + θ 2 w 2−α σxx |ξ1 | g , g + θ 2 w 2−α σx g , K¯ g 2 1
+ θ 2 w 2−α σx g , h¯ + θ 2 w 2−α σx |ξ1 | g 0 , g 0 + .
(4.10)
2
In the following, we only analyze some terms on the right side of the above equation since others are simple. Since 1
− 12
w −α σx2 K¯ σx
is a bounded operator from L 2ξ to L 2ξ , we have
1 1 −1 θ 2 w 2−α σx g , K¯ g θ 2 w −α σx2 g w −α σx2 K¯ σx 2 g 1 2 1 2 μθ σx2 w −α g + c σx2 g 1 2 μθ σx2 w −α g + c ξ1 a0 , a0 + + h¯ 22
(4.11)
where μ is a small constant. Since |σxx ξ1 | δ σx ν , we have
1 2
θ 2 w 2−α σxx |ξ1 | g , g δ θ 2 w 2−α νσx g , g .
(4.12)
What follows, we will estimate the last term of the right side of the above equality (4.10). When small, we have
1 2
θ 2 w 2−α σx ξ1 g 0 , g 0
= +
1
2 ξ 1 >0
c
2 θ 2 w 2−α σx ξ1 g 0 dξ
2 12 − σ (0,ξ ) −1 ξ M ∞ e θ 2 w 2−α σx ξ1 |a0 | + sM ∞2 M ω e σ (0,ξ ) g 0, ξ d ξ d ξ 1
ξ 1 >0
θ 2 w 2−α σx ξ1 |a0 |2 dξ + c
c
and s are sufficiently
ξ 1 >0
ξ1 <0
−1 2 2 σ (0,ξ ) ξ1 M ∞ ξ θ 2 w − 2 α σ x ξ1 s 2 M ∞ Mω e
ξ 1 >0 ξ <0
1 −1 2
w −2α ξ σx 0, ξ dξ dξ w −α σx ξ1 g 0 , g 0 − ×e
1 c θ 2 w 2−α σx ξ1 |a0 |2 dξ + w 2−α σx ξ1 g 0 , g 0 − .
−2 σ (0,ξ )
4
ξ 1 >0
(4.13)
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
735
By letting θ → 1, the following inequality 1 12 σx w −α ν 12 g c |ξ1 | 12 w −α a0 + h¯ 2 + w −α σx2 h¯ +
(4.14)
holds. From (4.2) and the above inequality, we can get
2
2 1 1 12 σx w 1 ν − 12 g x 2 σx2 w 1 σx − ν (ξ ) g + σx2 w 1 1 (h¯ + K¯ g ) ξ1 ξ1 1 1 2 1 2 1 2 c σx2 ν 2 g + σx2 g + σx2 h¯ .
(4.15)
2
Combining (4.14) with (4.15) completes the proof.
After the above preparations, we can estimate the g σ β now. Lemma 4.4. For 0 < α < 12 , β > 32 , the solution to the problem (3.6) satisfies 1 1
g σ β c ν −1 h¯ σ β + h¯ 2 + w −α σx2 h¯ + a0 +,β + |ξ1 | 2 w −α a0 +
(4.16)
where a0 +,β = supξ ∈ R 3 ,ξ1 >0 (1 + |ξ |)β |a0 (ξ )|. Proof. From the formula g = a˜ + U ( K¯ g + h¯ ) and by using the properties of U and K¯ , we have
g σ β c ˜a σ β + ν −1 K¯ g σ β + ν −1 h¯ σ β c ˜a σ β + g σ (β−1) + ν −1 h¯ σ β .
(4.17)
Notice that
12 − σ (0,ξ ) − 12
σ (0,ξ ) exp −κ (x, ξ ) sM ∞ M e M e g 0 , ξ d ξ ξ ω ∞ 1 ξ1 <0
σβ
1 c |ξ1 | 2 g 0 − .
(4.18)
Then, inserting the above result into the estimate (4.17), the following
1
g σ β c a0 +,β + g σ (β−1) + ν −1 h¯ σ β + |ξ1 | 2 g 0 − 1 c a0 +,β + g σ 0 + ν −1 h¯ + |ξ1 | 2 g 0
(4.19)
−
σβ
holds. In order to estimate g σ 0 , by using the result of Lemma 4.3 and Theorem 3.3, we have for
12 σx w γ g 2∞
L x ( L 2ξ )
1 4
< γ < 12 ,
2 1 σx2 w γ g L 2 ( L ∞ ) ξ
x
1 1 1 1 1 c σx2 w −(1−2γ ) ν 2 g + σx2 w 1 ν − 2 g x + σx2 g 1 1 c |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ .
(4.20)
Using the expression (4.3), we have 1
1
1
− 12
σx2 g = σx2 a˜ + σx2 U σx
1
1
σx2 K¯ g + σx2 h¯ .
(4.21)
From Theorem 3.3, Lemma 4.1 and Lemma 4.2, we have
12 σ x g
L 2ξ ( L ∞ x )
1 1 c ˜a L 2 ( L ∞ + ν −1 σx2 K¯ g L 2 ( L ∞ ) + ν −1 σx2 h¯ L 2 ( L ∞ ) x )
ξ
ξ
x
ξ
x
1 1 1 1 −1 c a0 + + |ξ1 | 2 g 0 − + σx2 K¯ σx 2 w −γ σx2 w γ g L 2 ( L ∞ ) + ν −1 σx2 h¯ L 2 ( L ∞ )
ξ
x
ξ
x
1 1 1 c a0 + + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ + ν −1 σx2 h¯ L 2 ( L ∞ ) .
ξ
x
(4.22)
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J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
Similarly, we have 1 1 1
g σ 0 c a0 +,0 + |ξ1 | 2 g 0 − + ν −1 σx2 K¯ g L ∞ + ν −1 σx2 h¯ L ∞ x,ξ
x,ξ
1 1 1 c a0 +,0 + |ξ1 | 2 g 0 − + σx2 g L 2 ( L ∞ ) + ν −1 σx2 h¯ L ∞
ξ
x
x,ξ
1 1 c a0 + + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ 1 1 + ν −1 σx2 h¯ 2 ∞ + ν −1 σx2 h¯ .
(4.23)
σ0
Lξ (L x )
Finally, when β > 32 , we have 1 1
g σ ,β c a0 +,β + a0 + + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ 1 1 + ν −1 σx2 h¯ L 2 ( L ∞ ) + ν −1 σx2 h¯ σ 0 + ν −1 h¯ σ β
ξ
c a0 +,β This completes the proof.
x
1 1 + |ξ1 | 2 w −(1−2γ )a0 + + h¯ 2 + w −(1−2γ ) σx2 h¯ + ν −1 h¯ σ β .
(4.24)
2
After the above estimates about the solution to the linearized Boltzmann equation with damping, we will prove the existence of a solution of the nonlinear Boltzmann equation with damping. Below, we will introduce the following two lemmas on the properties of Γ ( g , g ), see [5]. Lemma 4.5. The projection of Γ ( g , g ) on the null space of L vanishes and there exists a positive constant c, such that
−1 σ − σ ν e Γ e g , e − σ g
2
2 3−β1
g 2σ β σ β c exp − (δ x + l)
for any β >
1−β1 2
(4.25)
and sufficiently small constant > 0.
Lemma 4.6. When β >
5 2
and is sufficiently small, we have the results below:
− 12 σ − σ σ x e Γ e g , e − σ g c g 2σ β , σ − σ e Γ e g , e − σ g 2 c g 2σ β , 1 w −α σx2 e σ Γ e − σ g , e − σ g c g 2 , σβ
0<α<
1 2
(4.26)
.
In summary, we have the following existence theorem for the nonlinear problem with damping. −1
Theorem 4.7. When the boundary data F 0 satisfies that M ∞2 e σ (0,ξ ) ( F 0 − M ∞ + sM ω ξ <0 |ξ1 | M ∞ (0, ξ ) dξ ) is sufficiently small 1 in the norms · +,β , T ω < 2T ∞ , l is sufficiently large, , δ and s are sufficiently small and l−1 , then Eq. (3.5) has a unique solution F , such that e σ W 0−1 ( F − M ∞ ) is bounded in the norm · σ β , when β > 52 .
Proof. For any h which has · σ β < ∞, there exists a unique solution to the equation according to the theorem,
⎧ x > 0, ξ ∈ R 3 , ξ1 g x − σx ξ1 g − L g − D g = e σ Γ e − σ h, e − σ h , ⎪ ⎪ ⎨ 12 − σ (0,ξ ) −1 ξ M ∞ e g |x=0 = a0 (ξ ) + sM ∞2 M ω e σ (0,ξ ) g 0 , ξ d ξ , ξ1 > 0 . 1 ⎪ ⎪ ⎩
(4.27)
ξ 1 <0
Furthermore, from Lemmas 4.4 and 4.6, we will get
1
g σ β c h 2σ β + a0 +,β + |ξ1 | 2 w −α a0 + c h 2σ β + a0 +,β
(4.28)
since β > 52 . Hence, when a0 +,β is sufficiently small, there exists a solution g to (3.5) by contraction mapping theorem. Since g satisfies the inequality
J. Sun, Q. Tian / J. Math. Anal. Appl. 375 (2011) 725–737
737
Table 1
M∞ < − 1
− 1 < M∞ < 0 0 < M∞ < 1
M∞
>1
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 0
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 1
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 4
Codim({b0 ∈ L 2ξ1 ,+ | P 0+ ξ1 A γ (b0 ) = 0}) = 5
g σ β c g 2σ β + a0 +,β , the solution is unique.
2
As in [16], we will give an implicit solvability condition on the boundary data for the existence of a solution to the nonlinear Boltzmann equation. Let A γ be a nonlinear operator
A γ (b0 ) ≡ f (0, ·)
(4.29)
where f (x, ξ ) = e − σ (x,ξ ) g is given by
⎧ x > 0, ξ ∈ R 3 , ξ1 f x − L f = h − γ (δ x + l)−Θ P 0+ ξ1 f , ⎪ ⎪ ⎨ 12 −1 ξ M ∞ f 0, ξ dξ , ξ1 > 0. f |x=0 = b0 (ξ ) + sM ∞2 M ω 1 ⎪ ⎪ ⎩
(4.30)
ξ 1 <0
From the energy estimate on the solution to the nonlinear Boltzmann equation with damping, we know the operator satisfies
A γ : e σ (0,ξ ) b0 ∈ L 2ξ1 ,+ → A γ (b0 ) ∈ L 2|ξ | , and is bounded when b0 is sufficiently small. It is easy to see that the solution of (3.4) is a solution of (1.1) if P 0+ ξ1 A γ (b0 ) ≡ 0. Using the method in [16], we can get the key result in this paper. Theorem 4.8. The classification of the boundary data near the far field Maxwellian satisfying the solvability condition P 0+ ξ1 A γ (b0 ) ≡ 0 with respect to Mach number in the far field can be summarized in Table 1. References [1] K. Aoki, Y. Sone, T. Yamada, Numerical analysis of gas flows condensing on its plane condensed phase on the basis of kinetic theory, Phys. Fluids A 2 (1990) 1867–1878. [2] K. Aoki, K. Nishino, Y. Sone, H. Sugimoto, Numerical analysis of steady flows of a gas condensing on or evaporating from its plane condensed phase on the basis of kinetic theory: Effect of gas motion along the condensed phase, Phys. Fluids A 3 (1991) 2260–2275. [3] C. Bardos, R.E. Caflish, B. Nicolaenko, The Milne and Kramers problems for the Boltzmann equation of a hard sphere gas, Comm. Pure Appl. Math. 39 (1986) 323–352. [4] C. Cercignani, R. Illner, M. Pulvirenti, The Mathematical Theory of Dilute Gases, Appl. Math. Sci., Springer-Verlag, New York, 1994. [5] C.C. Chen, T.P. Liu, T. Yang, Exsistence of boundary layer solutions to the Boltzmann equation, Anal. Appl. 2 (4) (2004) 337–363. [6] F. Coron, F. Golse, C. Sulem, A classification of well-posed kinetic layer problems, Comm. Pure Appl. Math. 41 (1988) 409–435. [7] R.T. Glassey, The Cauchy Problem in Kinetic Theory, Society for Industrial and Applied Mathematics (SIAM), Philadephia, 1996. [8] F. Golse, B. Perthame, C. Sulem, On a boundary layer problem for the nonlinear Boltzmann equation, Arch. Ration. Mech. Anal. 103 (1) (1988) 81–96. [9] F. Golse, F. Poupaud, Stationary solutions of the linearized Boltzmann equation in a half-space, Math. Methods Appl. Sci. 11 (1989) 483–502. [10] H. Grad, Asymptotic theory of the Boltzmann equation, in: J.A. Laurmann (Ed.), Rarefied Gas Dynamics, vol. 1, Academic Press, New York, 1963, pp. 26– 59. [11] Y. Sone, Kinetic Theory and Fluid Dynamics, Birkhäuser, Berlin, 2002. [12] Y. Sone, Kinetic theory of evaporation and condensation linear and nonlinear problems, J. Phys. Soc. Japan 45 (1) (July 1978). [13] Q.Z. Tian, Existence of nonlinear boundary layer solution to the Boltzmann equation with physical boundary conditions, J. Math. Anal. Appl. 356 (1) (2009) 42–59. [14] Q.Z. Tian, J. Sun, Nonlinear stability of boundary layer solution to the Boltzmann equation with diffusive effect at the boundary, J. Math. Phys. 50 (2009) 103303 (SCI). [15] S. Ukai, K. Asano, On the Cauchy problem of the Boltzmann equation with a soft potential, Publ. Res. Inst. Math. Sci. 18 (2) (1982) 477–519. [16] S. Ukai, T. Yang, S.H. Yu, Nonlinear boundary layers of the Boltzmann equation: I. Existence, Comm. Math. Phys. 236 (3) (2003) 373–393. [17] S. Ukai, T. Yang, S.H. Yu, Nonlinear stability of boundary layers of the Boltzmann equation. I. The case M∞ < −1, Comm. Math. Phys. 244 (1) (2004) 99–109. [18] W.K. Wang, T. Yang, X.F. Yang, Existence of boundary layer to the Boltzmann equation with cutoff soft potentials, J. Math. Phys. 48 (7) (2007). [19] W.K. Wang, T. Yang, X.F. Yang, Nonlinear stability of boundary layers of the Boltzmann equation for cutoff hard potentials, J. Math. Phys. 47 (8) (2006). [20] X.F. Yang, Nonlinear stability of boundary layers for the Boltzmann equation with cutoff soft potentials, J. Math. Anal. Appl. 345 (2) (2008) 941–953.